Re: [其他] 數列一題

[LPH66 (杇瑣)]

※ 引述《coolbetter33 (香港3345678)》之銘言：
: 將一正整數列從頭算起.每三位便抽走.並重複此方法.問{1.2....1000}最後剩下哪三數
: 如:[1.2...13] =>[1.2.4.5.7.8.10.11.13] => [1.2.5.7.10.11] =>[1.2.7.10]
:  =>[1.2.10 ]
:   目前只確定會留下[1.2.?]  .....
計算一下每一輪剩下的數字個數

1000 -> 667 -> 445 -> 297 -> 198 -> 132
     -> 88 -> 59 -> 40 -> 27 -> 18
     -> 12 -> 8 -> 6 -> 4 -> 3

共計 15 輪

現在倒回去看  由觀察可知某一輪的第 k 個數字

在前一輪時它前面會多出 [(k-1)/2] 個數字  (這裡及以下 [] 都是高斯符號)

所以變成第 k + [(k-1)/2] 個數字

於是連續倒回去算:

#15  3              ← 代表 15 輪之後的第 3 個數字
#14  3 + [2/2] = 4  ← 變成 14 輪之後的第 4 個數字, 以下類推
#13  4 + [3/2] = 5
#12  5 + [4/2] = 7
#11  7 + [6/2] = 10
#10  10 + [9/2] = 14
#9   14 + [13/2] = 20
#8   20 + [19/2] = 29
#7   29 + [28/2] = 43
#6   43 + [42/2] = 64
#5   64 + [63/2] = 95
#4   95 + [94/2] = 142
#3   142 + [141/2] = 212
#2   212 + [211/2] = 317
#1   317 + [316/2] = 475
#0   475 + [474/2] = 712 ← 這就是第三個數了
                      (0 輪之後的第 712 個數即一開始的第 712 個數即 712 本身)

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